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\author{Dmitry Todorov}
\title{\bf \LARGE Normability of Metric Spaces of Keplerian Orbits}

\begin{document}
\maketitle

\begin{abstract}
\keywords{Keywords:} space of Keplerian orbits, topological space, normability
\end{abstract}

\section{Introduction}

In \cite{msko} were introduced spaces of curvilinear orbits $\Hb(b)$ and space of all orbits $\Hb.$ 
Both of these were merized and got the standart topologies from $\Rb^d,$ where $d$ equals $6$ and $7,$ respectively. Moreover, it has been shown, that both $\Hb(b)$ and $\Hb$ are 5-dimensional agebraic open manifolds without singular points, which are arcwise connected and, in addition, $\Hb$ is complete. These spaces are important and highly used, so it was a natural intention to introduce a richer structure of a normed space on it. This article shows that any attemt to do it will fail, because there are some topological arguments preventing these spaces from being normed.


\section{Definitions}
Consider the equation for $\vvec{r}\in \Rb^3:$
\begin{equation}
\ddot{\vvec{r}}+\varkappa^2 \frac{ \vvec{r} }{r^3}=0 \label{orbitode}
\end{equation}

\begin{deff}
We call an \emph{orbit} the set of points $\vvec{r}(t)\in\Rb^3$ for all possible $t,$ where is $\vvec{r}$ is some fixed solution of the equation \eqref{orbitode}.
\end{deff}

\noindent For $\Vc,\Ve\in \Rb^3,\ b\in[0,+\infty)$ consider equation and inequality
\begin{eqnarray} 
\Vc\cdot\Ve = 0 \label{Hbconecond} \\ 
c > b \label{Hboutsphcond}
\end{eqnarray}

The set of curvilinear orbits is defined as follows:
\begin{equation*}
\Hb(b)=\setdef{(\Vc,\Ve) \in \Rb^6}{\hbox{ $\Vc,\Ve$ satisfy  \eqref{Hbconecond} and $c$ satisfy \eqref{Hboutsphcond} } }
\end{equation*}


\noindent For $\Vc,\Ve\in \Rb^3,\ h\in\Rb$ consider equations
\begin{eqnarray} 
\Vc\cdot\Ve=0 \label{Hconecond} \\  
2hc^2-\varkappa^4(e^2-1)=0 \label{Hquadriccond}
\end{eqnarray}


The set of all orbits is defined as follows:
\begin{equation*}
\Hb=\setdef{(\Vc,\Ve,h) \in \Rb^7}{\hbox{ $\Vc,\Ve$ satisfy \eqref{Hconecond} and $\Vc,\Ve,h$ satisfy \eqref{Hquadriccond} } }
\end{equation*}

\section{Main Results}

\begin{theorem} There isn't exist a way to define a structure of a finite-dimensional topological linear space over the field $\Rb$ on the space $\Hb(b).$
\end{theorem}

\begin{theorem} There isn't exist a way to define a structure of a finite-dimensional topological linear space over the field $\Rb$ on the space $\Hb.$
\end{theorem}

\section{Proofs}

The key argument is that for a metric space of integer topological dimension to be normed (or at least have a structure op topological vector space) it is necessary to have very simple topological structure. For all topological facts used, see \cite{hatcher}.

At first notice, that both our space $\Hb(b)$ and $\Hb$ are $5$-dimensional. So, if there is exist a topological linear structure on one of them, such that it is consistent with metric topology, induced from ambient $\Rb^n,$ it should has the topological dimension equal to linear dimension (since finite dimensional topological vector space over $\Rb$ has topological dimension $n$ and manifolds of different topological dimensions cannot be homeomorphic). Here we omit the case when one of our orbit spaces can be made infinte-dimensional vector space, or vector space over field, different from $\Rb,$ but this cases is definitely out of our interest.

\subsection{Space of curvilinear orbits $\Hb(b)$}

\begin{statement} 
$\Hb(b)\cong TS^2 \times \Rb.$
\end{statement}
\begin{proof} At first, easy to see, that
$$\setdef{\Vc\in\Rb^3}{c>b} = \bigcup_{r>b} rS^2 = S^2 \times (b,+\infty)$$
For a fixed vector $\Vc$ equation \eqref{Hbconecond} defines a plane in $\Rb^3,$ orthogonal to $\Vc.$ Thus for a vectors from a sphere of a fixed radius $r$ we get tangent bundle $T(rS^2).$ The same could be done for every $r.$ It means $\Hb(b) = \bigcup\limits_{r>b} T(rS^2) \cong \bigcup\limits_{r>b} T(S^2) \cong TS^2 \times \Rb.$
%Consider basic rotation matrices $R_x(\theta),\ R_y(\theta),\ R_z(\theta)$, that rotate %vectors about the $x,y$ or $z$ axis, respectively. Let %$s:S^2\to(-\frac{\pi}{2},\frac{\pi}{2})\times [0,2\pi)$ be a transform from Cartesian to %spherical coordinates.

%\begin{equation*}
%(\frac{\Vc}{c},)
%\end{equation*}

\end{proof}

\begin{statement} 
$\Hb(b)$ is not homotopically equivalent to $\Rb^5.$
\end{statement}
\begin{proof} 
$TS^2$ is homotopically equivalent to $S^2$ (because every fiber bundle with contractible layers topologically equivalent to its base). Thus $\Hb(b)$ homotopically equivalent to $S^2 \times \Rb,$ which, in its turn homotopically equivalent to $S^2.$ But it is well known that $S^2$ isn't homotopically equivalent to $\Rb^5,$ because the latter has the property that any 2-sphere in it is homotopic to point (one can use a linear homotopy, for example), and $S^2$ hasn't. Strictly speaking this argument uses the simple fact that two homotopically equivalent space must have the same homotopy groups. In our case $\pi_2(S^2)=\Zb$ and all homotopy groups of $\Rb^5$ are trivial in all dimensions.
\end{proof}

\begin{corollary}[Theorem 1]
There isn't exist a way to define a structure of a finite-dimensional topological linear space over the field $\Rb$ on the space $\Hb(b).$
\end{corollary}
\begin{proof} 
Defining a topological linear structure on $\Hb(b)$ can be thought as finding a topological linear space, that is homeomorpic to $\Hb(b).$ If spaces are homeomorphic then they are homotopically equivalent, so the corollary follows immediately from previous statement.
\end{proof}

\subsection{Space of all orbits $\Hb$}

\begin{statement} 
$\Rb^5$ cannot be split by a compact set into two open connected subspaces, such that both of them have noncompact closure. I.e. complement of a compact set in $\Rb^5$ have exactly one connected component that has noncompact closure. 
\end{statement}
\begin{proof}
Any compact subset $M$ of the metric space is bounded, so there exist a ball $B$ of sufficiently large radius, such that $B$ contain $M.$ So any connected component of the complenet of $M$ conatin the complement of $M,$ which is connected itself and this means that there is only one connected component of the complenet of $M$ that contain the complement of $M.$ 
\end{proof}

\begin{statement} [Theorem 2]
$\Hb$ is not homeomorphic to $\Rb^5.$
\end{statement}
\begin{proof}
For $h=-1$ the equation \eqref{Hquadriccond} give an ellipsoid, which is compact. In $\Hb$ both two complements ($h<-1$ and $h>-1$, obviously both of them are connected) of this set would be unbounded, because for $h$ arbitrariliy big (positive or negative) there exist a nonepty set of solutions of the equations \eqref{Hquadriccond} and \eqref{Hconecond} (for example, one can take $\Vc=0,\ \Ve\in S^2$). 
\end{proof}

\begin{statement} 
$\Hb\cong \Rb \times S^2 \times S^2.$
\end{statement}
\begin{proof}
At first, we'll do some rescaling in equation \eqref{Hquadriccond} for it to reach the following form:
$$
tc^2+e^2=1
$$
In fact, we should divide $\Vc$ by $\sqrt{2/\varkappa^2}$ and take $t=-h.$ These actions don't affect equation \eqref{Hconecond}. We will show how to construct map $E_1:\Hb\to (0,\infty)\times S^2 \times S^2$ and its inverse. Let
\begin{eqnarray}
\vvec{u}=\vvec{c}+\vvec{e},\ \vvec{v}=\vvec{c}-\vvec{u} \label{uvdef} \\
\vvec{p}=\frac{\vvec{u}}{u},\ \vvec{q}=\frac{\vvec{v}}{v} \label{pqdef} \\
s=\begin{cases} - \displaystyle\frac{e^2}{c^2}, & c\neq 0 \\ -\infty, & c=0 \end{cases} \label{sdef}\\
k=\begin{cases} e^t-e^s, & c\neq 0 \\ e^t, & c=0 \end{cases} \label{kdef}
\end{eqnarray} 
We define $E_1(t,\vvec{c},\vvec{e})$ to be $(k,\vvec{p},\vvec{q}),$ where $k,p$ and $q$ are defined according to equalities \eqref{uvdef}-\eqref{kdef}. 

The inverse map $E_2$ may be constructed in a following way: for $\vvec{p},\vvec{q}\in S^2\subset\Rb^3$ take
\begin{eqnarray*}
a=1-\vvec{p}\cdot\vvec{q},\ b=1+\vvec{p}\cdot\vvec{q} \\
s= \begin{cases} -\displaystyle\frac{b}{a}, & \vvec{p}\neq \vvec{q} \\ -\infty, & \vvec{p} = \vvec{q} \end{cases}
\end{eqnarray*} 
Then find $t$ from the equation \eqref{kdef} and take 
$$
r=\sqrt{\frac{2}{at+b}}
$$
Finally, one should find $\vvec{c}$ and $\vvec{e}$ from equations \eqref{uvdef} for $\vvec{u}=r\vvec{p},\ \vvec{v}=r\vvec{q}$ and put $E_2(k,\vvec{p},\vvec{q})=(t,\vvec{c},\vvec{e}).$
 

\end{proof}

\begin{thebibliography}{99}
\bibitem{msko} Konstantin V. Kholshevnikov, Metric Spaces of Keplerian Orbits, Celestial Mechanics and Dynamical Astronomy, Volume 100, Issue 3, pp. 169-179
\bibitem{hatcher} Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002
\end{thebibliography}

\end{document}
